0
$\begingroup$

I have the following system \begin{gathered} \frac{{dx}}{{dt}} = A*cos(y) + B*cos (b*t) \hfill \\ \frac{{dz}}{{dt}} = A*sin(y) - Ccos (c*t) \hfill \\ \frac{{dy}}{{dt}} = D*tan(d*t), \hfill \\ \end{gathered}

where $$A,B,C,D,b,c,d$$ are constants.

$$t \in (0,T)$$, where $$T < \infty $$

I do not have any idea what approach to use to solve it

What I have done is as follows: 1) Integration of y $$y = D{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|_{t\max }} - D{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|_{t\min }}$$ 2)I inserted y into the equation with x $$x = A\int {\cos \left( {D{{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|}_{t\max }} - D{{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|}_{t\min }}} \right)} dt + smth\_simple$$

But this is very fastly oscillating function and limits of integral confuse me. Notice that $$\left| {t\max - t\min } \right| \leqslant 10\sec $$ P.S. y can change only within 10 sec period, then it is constant

Probably I should assign t to tmax . And if tmin is equal to 0. I can integrate x from 0 to something. But what if tmin is not zero?

$\endgroup$
6
  • $\begingroup$ I'll edit my post to show $\endgroup$
    – Ruslan1935
    Aug 1 '17 at 18:42
  • $\begingroup$ And what is $g$? $\endgroup$ Aug 1 '17 at 18:52
  • $\begingroup$ tg a.k.a. tan a.k.a. Tangent function $\endgroup$
    – Ruslan1935
    Aug 1 '17 at 18:59
  • $\begingroup$ You mean the following? $$\dfrac{dy}{dt} = D \tan (d \times t)$$ $\endgroup$
    – Moo
    Aug 1 '17 at 19:12
  • $\begingroup$ Yes, this is what I mean $\endgroup$
    – Ruslan1935
    Aug 1 '17 at 19:15
0
$\begingroup$

If $g(t)$ is a known function, integrate the equation for $dy/dt$ to get $y$, then substitute the result into the equations for $dx/dt$ and $dz/dt$ and integrate.

$\endgroup$
1
  • $\begingroup$ I have added new information to my question $\endgroup$
    – Ruslan1935
    Aug 1 '17 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.